Problem: $f\,^{\prime}(x)=12x^2-6x+2$ and $f(-1)=3$. $f(2) = $
Explanation: Finding $f(x)$ We have $f'(x)=12x^2-6x+2$ and we want to find $f(x)$ : $\begin{aligned}f(x) &= \int f'(x)\,dx \\\\ & = \int (12x^2-6x+2)\,dx \\\\ & = {4x^3-3x^2+2x} {+ C} \end{aligned}$ Finding $ C$ Goal: We need to find $ C$ such that $f(-1)=3$. Here's what we get when we plug in $-1$ : $\begin{aligned}f(-1)&={4(-1)^3-3(-1)^2+2(-1)} {+ C}\\\\ &={-9} {+ C} \end{aligned}$ We are given that this must equal $3$ : $3 = {-9} {+ C}$ Solving the equation gives us ${C=12}$. Finding $f(2)$ Now, we have that $f(x)={4x^3-3x^2+2x} {+12}$. Let's find $f(2)$ by plugging in $2$ : $\begin{aligned}f(2)&=4(2)^3-3(2)^2+2(2) + 12\\\\ &=36 \end{aligned}$ The answer $f(2) = 36$